BASIC CAL PROJECT MAGAZINE (1)

VOLUME 1 FEBRUARY 2023 The Math Kenneth Lasala andWiitszalrimd its

ABOUT THE COVER In partial fulfillment of our course requirement for Basic Calculus 11, we present to you our exclusive magazine covering topics from the quarter featuring the Math Wizard himself, Kenneth Lasala

CBLAS

CONTENTS 5 6 Introduction to limits Intermediate Value Theorem 8 9 Extreme Value Theorem Continuity 11 13 Discontinuity Slope of a Secant Line 15 Slope of a Tangent Line

INTRODUCTION TO LIMITS Consider a function f of a single variable x. Consider a constant c which the variable x will approach ( may or may not in the domain of f ) . The limit, to be denoted by L, is the unique real value that f(x) will approach x as approaches c. In symbols, we write this process as;

Limits are the backbone of calculus, and calculus is called the Mathematics of Change. The study of limits is necessary in studying change in detail. The study of limits is necessary in studying change in detail. The evaluation of a particular limit is what underlies the formulation of the derivative and antiderivative (indefinite integral) of a function.

Step 2 : Substitue theMEDIATEIVT value of f(c) to the f(x) If a function f(x) is to test the IVT. continuous over a closed interval [a, b], then for Final Answer : SinceTHEOREevery value m between 3/2 oe 1.5 is within∈f(a) and f(b), there is a [1,5], then IVT holds value c [a, b] such true when f(c) is -2 that f(c) = m Consider the function f(x) = 2x-5 on the closed TERMEinterval [1, 5]. Show that the intermediate value theorem holds E THEOtrueiff(c)=-2 Step 1 : substitute the x variable of the function with the following interval to get the y MTEHDEIAOTRintervals.

evt extreIf a function f(x) is continuous over a closed interval [a,b], theorthen f(x) is guaranteed to reach a maximum and a roermeme vminimumon[a,b]. rmeme v(eav

We observe that lim f (x) f (c) if the function is x -> c not continuous at point c. A function f is continuous when, for every value c in its domain: f (c) is defined, and lim f (x) = f (c) x -> c THREE CONDITIONS OF CONTINUITY 1. f (x) exists 2.lim f (x) exists x -> c 3. f (c) = lim f (x) x -> c

An interval includes all the numbers that comes between two particular numbers. This range includes all the real numbers between those two numbers. HOW IS A FUNCTION CONTINUOUS AT AN INTERVAL? A function is continuous at an interval if the values included in the interval are all part of the graph or exists as a solution of the function.

The function of the graph which is not DIS connected with each CON other is known as a discontinuous function. A function f(x) is said to TI have a discontinuity of the first kind at x = a, if the left-hand limit of NUITY f(x) and right-hand limit of f(x) both exist but are not equal. REMOVABLE The discontinuity occurs when there a is a hole in the graph of the function. The function can be redefined to remove the discontinuity. Happens in indeterminate functions.

This discontinuity occurs JUMPwhen the graph of the function stops at one point and seems DISCOto jump at another point. Both left hand and right-hand NTINlimit exists but are not equal. UITY At least one of the a two limits are infinite. Has a vertical INFINI asymptote. TE Happens in DISCO NTINU rational, logarithmic, and trigonometric function. a ITY

SLOPE Give any two points on the line we can find the slope by finding the change in y over the change in x. SLOPE INTERCEPT FORM Slope can also be determined given the form y=mx+b wherein m is the slope and b is the y intercept. POINT SLOPE FORM We can write the equation of the line given the slope and any point SLOPEfrom the line by using the point- OF Aslope form. TANGEN T LINE

SLOPE SECANT LINE OF AIs the line that passes through two TANGENpoints of a curve. T LINE SLOPE OF THE SECANT LINE m= f (x) - f (a) x-a

TANGENT LINE Is a straight line that touches a curve at a single point which we call the point of tangency. Slope of a tangent line Step 1 : to find the lxi-m> c= f (x) - f (a) value of f(a) you just x-a need to substitute the value of x (x is also your a) to the f(x). Step 2 : use the formula Step 4 : combined like to find the slope of the terms in the numerator. tangent line. Step 3 : Since there is a Step 5 : we just need to radical, we need to use multiply the expression to -1 the conjugate pair. to have a positive variable. Step 6 : factor the numerator. Step 7 : cancel the similar expression. Step 8 : once you have your new expression. Change your x with the value of your c to find the limit/slope. Step 9 : simplify your answer if possible.

Example : f(x) = √25 - x^2 a = 4 f(a) = 3 xli-m> c= f (x) - f (a) x-a lxi-m> 4= √25-x^2 - 3 √25-x^2 + 3 x-4 √25-x^2 + 3 xli-m> 4= (x 25-x^2 - 9 + 3) - 4) (√25-x^2 xli-m> 4= (x - x^2-16 + 3) 4) (√25-x^2 xli-m> 4= (x - (x-4) (x+4) + 3) 4) (√25-x^2 xli-m> 4= (x+4) (√25-x^2 + 3) xli-m> 4= (4+4) (√25-(4)^2 + 3) 8 4 m= 6 3

TANGENT LINE Is a straight line that touches a curve at a single point which we call the point of tangency. Equation of a Tangent Line Step 1 : after solving y - f(a)= m (x-a) for the slope, use the formula to find the Step 2 : distribute the m equation of a tangent to your (x-a) line. Step 3 : transpose -3 to the other side to isolate the value y. Note: change the sign as you transpose Step 4 : find the difference between -16/3 and +3

Example : a=4 f(a) = 3 m = 8 4 f(x) = √25 - x^2 3 6 y - f(a)= m (x-a) y-3= 4 (x - 4) 3 y-3= x -4 16 33 y = 4 x - 16 +3 33 y = 4 x - 16 + 9 3 33 y = 4 x- 7 3 3



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